# Find value of (n^1 + n^2 + n^3 + n^4) mod 5 for given n

You are given a function f(n) = (n1 + n2 + n3 + n4), you have to find the value of f(n) mod 5 for any given value of positive integer n.
Note: n may be large enough, such that f(n) > 1018.
Examples :

```Input : n = 4
Output : 0
Explanation : f(4) = 4 + 16 + 64 + 256 = 330,
f(4) mod 5 = 330 mod 5 = 0.

Input : n = 1
Output : 4
Explanation : f(1) = 1 + 1 + 1 + 1 = 4,
f(1) mod 5 = 4.
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

First of all for solving this approach you may find the value of (n1 + n2 + n3 + n4) mod 5 directly with the help of anypower function and modulo operator.
But For the larger value of n, your result will be wrong because for large n value of f(n) may go out of range from long long int in that case you have to opt some other efficient way.
To solve this question lets do some small mathematical derivation for f(n).

```f(n) = (n1 + n2 + n3 + n4)
= (n) (n+1) (n2+1)
Now, for finding f(n) mod 5 we must take care of unit digit of f(n) only,
also as f(n) mod 5 is dependent on n%5, (n+1)%5 & (n2+1)%5,
if any of these three result in zero then our whole result is 0.
So, if n = 5, 10, .. 5k then n mod 5 = 0 hence f(n) mod 5 = 0.
if n = 4, 9, .., (5k-1) then (n+1) mod 5 = 0 hence f(n) mod 5 = 0.
if n = 3, 8, 13..., (5k-2) f(n) mod 5 = (3 * 4 * 10) mod 5 = 0
if n = 2, 7, 12..., (5k-3) f(n) mod 5 = (2 * 3 * 5) mod 5 = 0.
if n = 1, 6, 11..., (5k-4) f(n) mod 5 = (1 * 2 * 2) mod 5 = 4.
```

After above analysis we can see that if n is of form 5k+1 or say 5k-4 then f(n) mod 5 = 4, other wise f(n) = 0.
I.E. if(n%5 == 1 ) result = 4,
else result = 0.

## C++

 `// finding the value of f(n) mod 5 for given n. ` `#include ` `using` `namespace` `std; ` ` `  `// function for f(n) mod 5 ` `int` `fnMod(``int` `n) ` `{ ` `    ``// if n % 5 == 1 return 4 ` `    ``if` `(n % 5 == 1) ` `        ``return` `4; ` ` `  `    ``// else return 0 ` `    ``else` `        ``return` `0; ` `} ` ` `  `// driver program ` `int` `main() ` `{ ` `    ``int` `n = 10; ` `    ``cout << fnMod(n) << endl; ` `    ``n = 11; ` `    ``cout << fnMod(n) << endl; ` `    ``return` `0; ` `} `

## Java

 `// Java code to finding the value  ` `// of f(n) mod 5 for given n. ` `import` `java.io.*; ` ` `  `class` `GFG  ` `{ ` `    ``// function for f(n) mod 5 ` `    ``static` `int` `fnMod(``int` `n) ` `    ``{ ` `        ``// if n % 5 == 1 return 4 ` `        ``if` `(n % ``5` `== ``1``) ` `            ``return` `4``; ` `     `  `        ``// else return 0 ` `        ``else` `            ``return` `0``; ` `    ``} ` `     `  `    ``// Driver program ` `    ``public` `static` `void` `main (String[] args) ` `    ``{ ` `        ``int` `n = ``10``; ` `        ``System.out.println(fnMod(n)); ` `        ``n = ``11``; ` `        ``System.out.println(fnMod(n)); ` ` `  `    ``} ` `} ` ` `  `// This code is contributed by vt_m. `

## Python3

 `# Python3 program to find the value ` `# of f(n) mod 5 for given n. ` ` `  `# Function for f(n) mod 5 ` `def` `fnMod(n): ` ` `  `    ``# if n % 5 == 1 return 4 ` `    ``if` `(n ``%` `5` `=``=` `1``): ` `        ``return` `4` ` `  `    ``# else return 0 ` `    ``else``: ` `        ``return` `0` ` `  `# Driver Code ` `n ``=` `10` `print``(fnMod(n)) ` ` `  `n ``=` `11` `print``(fnMod(n)) ` ` `  `# This code is contributed by Smitha Dinesh Semwal `

## C#

 `// Code for finding the value ` `// of f(n) mod 5 for given n. ` `using` `System; ` ` `  `class` `GFG { ` `    ``// function for f(n) mod 5 ` `    ``static` `int` `fnMod(``int` `n) ` `    ``{ ` `        ``// if n % 5 == 1 return 4 ` `        ``if` `(n % 5 == 1) ` `            ``return` `4; ` ` `  `        ``// else return 0 ` `        ``else` `            ``return` `0; ` `    ``} ` ` `  `    ``// Driver program ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int` `n = 10; ` `        ``Console.WriteLine(fnMod(n)); ` `        ``n = 11; ` `        ``Console.WriteLine(fnMod(n)); ` `    ``} ` `} ` ` `  `// This code is contributed by vt_m. `

## PHP

 ` `

Output :

```0
4
```

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Improved By : vt_m